Mathematics (Core)
Paper 1 | Objectives | 50 Questions
WASSCE/WAEC MAY/JUNE
Year: 2018
Level: SHS
Time:
Type: Question Paper
Answers provided
FREE
No description provided
Feedbacks
This paper is yet to be rated
Paper 1 | Objectives | 50 Questions
WASSCE/WAEC MAY/JUNE
Year: 2018
Level: SHS
Time:
Type: Question Paper
Answers provided
No description provided
This paper is yet to be rated
Did you apply the MTN Ghana Bright Scholarships 2022 for Ghana Undergraduate Students?
Good jobs available to people without a college degree, how to get good job without a college degree?
These are the best study techniques and methods that get higher grades in any school tests or exams.
| # | Question | Ans |
|---|---|---|
| 1. |
Simplify: \(\sqrt{108} + \sqrt{125} - \sqrt{75}\) A. \(\sqrt{3} + 5\sqrt{5}\) B. \(6 \sqrt{3} - 5 \sqrt{5}\) C. \(6 \sqrt{3} + \sqrt{2}\) D. \(6\sqrt{3} - \sqrt{2}\)
Show Content
Detailed Solution\(\sqrt{108} + \sqrt{125} - \sqrt{75}\)= \(\sqrt{3 \times 36} + \sqrt{5 \times 25} - \sqrt{3 \times 25}\) = \(6 \sqrt{3} + 5 \sqrt{5} - 5 \sqrt{3}\) = \(\sqrt{3} + 5\sqrt{5}\) |
|
| 2. |
Evaluate: \((64^{\frac{1}{2}} + 125^{\frac{1}{3}})^2\) A. 121 B. 144 C. 169 D. 196
Show Content
Detailed Solution\([64^{\frac{1}{2}} + 125^{\frac{1}{3}}]^2\) = \([\sqrt{64} + \sqrt[3] {125}]^2\)\([8 + 5]^2\) = \([13]^2\) = 169 |
|
| 3. |
Given that y varies inversely as the square of x. If x = 3 when y = 100, find the equation connecting x and y. A. \(yx^2 = 300\) B. \(yx^2 = 900\) C. y = \(\frac{100x}{9}\) D. \(y = 900x^2\)
Show Content
Detailed SolutionY \(\alpha \frac{1}{x^2} \rightarrow y = \frac{k}{x^2}\)If x = 3 and y = 100, then, \(\frac{100}{1} = \frac{k}{3^2}\) \(\frac{100}{1} = \frac{k}{9}\) k = 100 x 9 = 900 Substitute 900 for k in y = \(\frac{k}{x^2}\); y = \(\frac{900}{x^2}\) = \(yx^2 = 900\) |
|
| 4. |
Find the value of x for which \(32_{four} = 22_x\) A. three B. five C. six D. seven
Show Content
Detailed Solution\(32_4 = 22_x\)\(3 \times 4^1 + 2 \times 4^o\) = \(2 \times x^1 + 2 \times x^o\) 12 + 2 x 1 = 2x + 2 x 1 14 = 2x + 2 14 - 2 = 2x 12 = 2x x = \(\frac{12}{2}\) x = 6 |
|
| 5. |
Simplify; 2\(\frac{1}{4} \times 3\frac{1}{2} \div 4 \frac{3}{8}\) A. \(\frac{5}{9}\) B. 1\(\frac{1}{5}\) C. 1\(\frac{1}{4}\) D. 1\(\frac{4}{5}\) |
|
| 6. |
There are 250 boys and 150 girls in a school, if 60% of the boys and 40% of the girls play football, what percentage of the school play football? A. 40.0% B. 42.2% C. 50.0% D. 52.5%
Show Content
Detailed SolutionPopulation of school = 250 + 150 = 40060% of 250 = \(\frac{\text{60%}}{\text{100%}}\) x 250 = 150 40% of 150 = \(\frac{\text{40%}}{\text{100%}}\) x 150 = 60 Total number of students who plays football; 150 + 60 = 210 Percentage of school that play football; \(\frac{210}{400}\) x 100% = 52.5% |
|
| 7. |
If \(\log_{10}\)(6x - 4) - \(\log_{10}\)2 = 1, solve for x. A. 2 B. 3 C. 4 D. 5
Show Content
Detailed Solution\(\log_{10}\)(6x - 4) - \(\log_{10}\)2 = 1\(\log_{10}\)(6x - 4) - \(\log_{10}\)2 = \(\log_{10}\)10 \(\log_{10}\)\(\frac{6x - 4}{2}\) - \(\log_{10}\)10 \(\frac{6x - 4}{2}\) = 10 6x - 4 = 2 x 10 = 20 6x = 20 + 4 6x = 20 x = \(\frac{24}{6}\) x = 4 |
|
| 8. |
If F = \(\frac{9}{5}\)C + 32, find C when F = 98.6 A. 30 B. 37 C. 39 D. 41
Show Content
Detailed SolutionF = \(\frac{9}{5}\)C + 32When F = 98.6 98.6 = \(\frac{9}{5}\)C + 32 98.6 - 32 = \(\frac{9}{5}\)C 66.6= \(\frac{9}{5}\)C 66.6 x 5 = 9C C = \(\frac{66.6 \times 5}{9}\) = 37 |
|
| 9. |
If y + 2x = 4 and y - 3x = -1, find the value of (x + y) A. 3 B. 2 C. 1 D. -1
Show Content
Detailed Solutiony + 2x = 4 .....(1)y - 3x = -1 ......(2) Subtract (2) from (1) 2x - (-3x) = 4 - (-1) 2x + 3x = 4 + 1 5x = 5 X = \(\frac{5}{5}\) = 1 Substitute 1 for x in (1); y + 2(1) = 4 y + 2 = 4 y = 4 - 2 = 2 Hence, (x + y) = (1 + 2) = 3 |
|
| 10. |
If x : y : z = 3 : 3 : 4, evaluate \(\frac{9x + 3y}{6z - 2y}\) A. 1\(\frac{1}{2}\) B. 2 C. 2\(\frac{1}{2}\) D. 3
Show Content
Detailed SolutionIf x : y : z = 3 : 3 : 4, evaluate \(\frac{9x + 3y}{6x - 2y}\)\(\frac{x}{y}\) = \(\frac{2}{3}\) and \(\frac{y}{z}\) = \(\frac{3}{4}\) Thus; x = \(\frac{2}{3}T_1\) and z = \(\frac{3}{5}T_1\) y = \(\frac{3}{7}T_2\) and z = \(\frac{4}{7}T_2\) Using y = y \(\frac{3}{5}T_1\) = \(\frac{3}{7}T_2\); \(\frac{T_1}{T_2}\) = \(\frac{3}{7}\) x \(\frac{5}{3}\) \(\frac{T_1}{T_2}\) = \(\frac{15}{21}\) \(T_1\) = 15 and \(T_2\) = 21 Therefore; x = \(\frac{2}{5}\) x 15 = 6 y = \(\frac{3}{5}\) x 15 = 9 y = \(\frac{3}{7}\) x 21 = 9 (again) z = \(\frac{4}{7}\) x 21 = 12 Hence; \(\frac{9x + 3y}{6z - 2y}\) = \(\frac{9(6) + 3(9)}{6(12) - |
Preview displays only 10 out of the 50 Questions